[Math] How to find minimum n for Composite Trapezoidal rule

integrationnumerical methods

I'm taking a course on numerical methods to be able to program math, it's part of the game option of my compsci bachelors.This is part of one of the questions:

Given the integral: $\int_6^{12} (9x^3 – x^4)dx$

A. Using the Composite Trapezoidal Rule, what is the min. n required to have an accuracy of six decimal places (TOL. $\le 10^{-6}$), in the approximation of the given integral. n must be solved theoretically.

B. just use the composite trapezoidal rule with said n, which is easy since I programmed the rule (code here: http://pastebin.com/QzRdijrh)

C. Same as A but for Composite Simpson's rule.

D. Same as B but for Composite Simpson's rule (code here: http://pastebin.com/V9DYdRXU)

So the problem I'm having is I can't remember how to find minimum n, I know for Bisection Algorithm you make $2^n \le (b-a)/TOL$ But I forgot how to apply that to other rules…

Best Answer

The error term of the composite trapezoidal rule is $\frac {(b-a)^3}{12n^2}f''(\xi)$ so you need to evaluate the second derivative of your function, find its maximum and plug everything in. You will get something like error $\lt \frac a{n^2}$. Then you choose $n$ to get the error lower than $10^{-6}$. For the composite Simpson's rule, the approach is the same, but you have a different expression for the error.

Related Solutions

[Math] Finding $n$ value for trapezoid and midpoint rule errors

What are the given correct answers?

Your work seems correct. But of course $n$ is a positive integer, so you need to round your values up to the ceiling of your calculated answers. So $n=248$ guarantees the desired precision for the trapezoidal rule and $n=175$ does so for the midpoint rule.

However, those values were gotten by using the formulas that guarantee the desired precision, given a bound on the absolute value of the second derivative. You may get much better precision than the guaranteed value. Using $n=175$ I get the error $0.000028237574$ using the midpoint rule, much better than the desired $0.001$. You can get the desired error with much smaller values of $n$.

You should repeat the calculations of the approximate integrals to find just which value of $n$ actually gives you the desired error. For the midpoint rule in your situation, I get $n=93$ as the smallest $n$, which give the error $0.000099982028$. ($n=92$ gives $0.000102167251$.) For the trapezoidal rule I get $n=132$, giving the error $0.000099262595$.

Is the given correct answer for the midpoint rule $n=175$, my theoretical answer, or $n=93$, my calculated answer, or something else?

[Math] Numerical integration in Matlab (Simpson’s rule)

A couple of notes. Since the integrand is never less than $1$, you know that $x_l<170$. In fact, when $x=170$, the integrand will be about $1.087$, so the eventual answer is bounded below by $\frac{170}{1.087}=156.5$. That gives you a pretty good idea about the starting point.

Bisection is useful when you don't have the derivative available, but here the you have the derivative in the form of the integrand, so it seems to be simpler to implement Newton's method in this case unless bisection is the point of the assignment.

Also you know the integral up to a point near to the actual root at each step after the first, so you need not integrate all the way from zero to the next guess for the root, just from the previous guess. Even if you integrated from zero every time, this program would zip through pretty fast anyway.

EDIT: Newton's method

% simp.m

f = @(x) sqrt(1+(x^2/68000)^2);

err = 1;
tol = 1.0e-8;
N = 10000;
a = 0;
b = 163;
while abs(err) > tol,
    h = (b-a)/N;
    y = f(a);
    for i = 1:N/2,
        y = y+4*f((2*i-1)*h)+2*f(2*i*h);
    end
    y = y+4*f(b-h)+f(b);
    y = h/3*y-170;
    yp = f(b);
    err = y/yp;
    b = b-err;
end
b

Best Answer

Related Solutions

[Math] Finding $n$ value for trapezoid and midpoint rule errors

[Math] Numerical integration in Matlab (Simpson’s rule)

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